A ring of mass M and radius R is rotating with angular speed omega about a fixed vertical axis passi

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6.System of Particles and Rotational Motion

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A ring of mass $M$ and radius $R$ is rotating with angular speed $\omega$ about a fixed vertical axis passing through its centre $O$ with two point masses each of mass $\frac{ M }{8}$ at rest at $O$. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is $\frac{8}{9} \omega$ and one of the masses is at a distance of $\frac{3}{5} R$ from $O$. At this instant the distance of the other mass from $O$ is

$\frac{2}{3} R$

$\frac{1}{3} R$

$\frac{3}{5} R$

$\frac{4}{5} R$

(IIT-2015)

Solution

Using conservation of angular momentum

$mR ^2 \omega=\left(m R^2 \times \frac{8 \omega}{9}\right)+\left(\frac{ m }{8} \times \frac{9 R^2}{25} \times \frac{8 \omega}{9}\right)+\left(\frac{ m }{8} \times x^2 \times \frac{8 \omega}{9}\right) \Rightarrow x=\frac{4 R}{5}$

Standard 11

Physics

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$(A)$ $v=\sqrt{\frac{k}{2 m}} R$

$(B)$ $v=\sqrt{\frac{k}{m}} R$

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$(D)$ $\mathrm{L}=\sqrt{\frac{\mathrm{mk}}{2}} \mathrm{R}^2$

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(IIT-2018)

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(JEE MAIN-2024)

A fan of moment of inertia $0.6\,kg \times m^2$ is turned upto a working speed of $0.5$ revolutions per second. The angular momentum of the fan is

easy

Solution

Similar Questions

A hemisphere of radius $R$ and of mass $4m$ is free to slide with its base on a smooth horizontal table. A particle of mass $m$ is placed on the top of the hemisphere. The angular velocity of the particle relative to hemisphere at an angular displacement $\theta $ when velocity of hemisphere $v$ is

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